Answer :

Answer:

No

Step-by-step explanation:

Given

f(x) = [tex]\sqrt{x-10}[/tex] + 11 ← substitute x = 1

f(1) = [tex]\sqrt{1-10}[/tex] + 11

     = [tex]\sqrt{-9}[/tex] + 11

[tex]\sqrt{-9}[/tex]  is not real and therefore x = 1 does not have a real output for f(x)

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